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We know that for a Hilbert triple $V \subset H \subset V^{'}$, if we have $u, v \in L^2(0,T;V)$ with $u',v' \in L^2(0,T;V')$ then $$\frac{d}{dt}(u(t), v(t))_H = u'(t)(v(t)) + v'(t)(u(t))$$ where the $'$ means the weak derivative.

Is there a formula involving weak derivatives for $$\frac{d}{dt}f(t)(u(t))$$ where $f(t) \in V^*$ and $u(t) \in V$?

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  • $\begingroup$ $u'(t)\in V^*$ and $v(t)\in H$. How do you define $v'(t)(\; v(t)\;)$? $\endgroup$
    – Liviu Nicolaescu
    Commented Jan 20, 2013 at 16:47
  • $\begingroup$ @Liviu Apologies, $v(t) \in V$. I edited. $\endgroup$
    – Chris
    Commented Jan 20, 2013 at 17:28
  • $\begingroup$ Why doesn't the usual product formula work? $\endgroup$
    – Deane Yang
    Commented Jan 20, 2013 at 17:30
  • $\begingroup$ @Deane by $f(t)(u(t))$ I mean $\langle f(t), u(t) \rangle_{V^*, V}$, the pairing. $\endgroup$
    – Chris
    Commented Jan 20, 2013 at 17:33
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    $\begingroup$ Why can't you use essentially the same definitions and proof used for the Hilbert space result? $\endgroup$
    – Deane Yang
    Commented Jan 20, 2013 at 18:51

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